The sum of the squares of two consecutive positive numbers is 61. What is the smaller number?

2 Answers
Jun 28, 2016

Smallest no is 5. Other no is 6.

Explanation:

Let the smaller no be n. So the other no is n+1

Given: n^2+(n+1)^2 = 61 =>

n^2 + n^2 + 2n + 1 - 61 = 0

n^2 + n^2 + 2n + 1 - 61 = 0

2n^2 + 2n - 60 = 0

n^2 + n - 30 = 0

(n+6)(n-5) = 0 => n = -6 or n=5#

Discard -6 as n is positive. So, n = 5

Jun 28, 2016

The smaller number is 5.

Explanation:

Let the smaller number be x and then next consecutive number is x+1. As the sum of the squares of these two consecutive positive numbers is 61, we have

x^2+(x+1)^2=61 or

x^2+x^2+2x+1=61 or

2x^2+2x-60=0 or (diving each side by 2)

x^2+x-30=0 or

x^2+6x-5x-30=0 or

x(x+6)-5(x+6)=0 or

(x+6)(x-5)=0

Hence x=-6 or x=5.

But as we need positive numbers only, x=5.